<p>Because of floating point imprecision, you're unlikely to get the value you expect from the <code>BigDecimal(double)</code> constructor. </p>
<p>From <a href="http://docs.oracle.com/javase/7/docs/api/java/math/BigDecimal.html#BigDecimal(double)">the JavaDocs</a>:</p>
<blockquote>The results of this constructor can be somewhat unpredictable. One might assume that writing new BigDecimal(0.1) in Java creates a BigDecimal which is exactly equal to 0.1 (an unscaled value of 1, with a scale of 1), but it is actually equal to 0.1000000000000000055511151231257827021181583404541015625. This is because 0.1 cannot be represented exactly as a double (or, for that matter, as a binary fraction of any finite length). Thus, the value that is being passed in to the constructor is not exactly equal to 0.1, appearances notwithstanding.</blockquote>
<p>Instead, you should use <code>BigDecimal.valueOf</code>, which uses a string under the covers to eliminate floating point rounding errors.</p>
<h2>Noncompliant Code Example</h2>

<pre>
double d = 1.1;

BigDecimal bd1 = new BigDecimal(d); // Noncompliant; see comment above
BigDecimal bd2 = new BigDecimal(1.1); // Noncompliant; same result
</pre>
<h2>Compliant Solution</h2>

<pre>
double d = 1.1;

BigDecimal bd1 = BigDecimal.valueOf(d);
BigDecimal bd2 = BigDecimal.valueOf(1.1);
</pre>
<h2>See</h2>

<ul>
<li> <a href="https://www.securecoding.cert.org/confluence/x/NQAVAg">CERT, NUM10-J</a> - Do not construct BigDecimal objects from floating-point literals</li>
</ul>

